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Math Help - Trig Intervals

  1. #1
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    Trig Intervals

    I'm trying to figure out this problem. I think I have the answer but I'm not sure.

    Find all solutions on the interval [0,2  {\pi} ]. When appropriate put you answer in interval notation.

     cos(x)-\frac{\sqrt{3}}{2}>0

    1) this is the same as  cos(x)+0>\frac{\sqrt{3}}{2} correct?
    2) I know that cosx is equal to  \frac{\sqrt{3}}{2} at  \frac{\pi}{6},\frac{11\pi}{6}

    but, since it is asking for greater than, I graphed out cosx and thought that the answer is 0,  {2\pi} . am I close?
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  2. #2
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    Hi discobob,

    What exactly does the graph of cos(x) look like from 0 to 360 degrees?

    It starts at cos(0)=1,
    drops to zero at 90 degrees,
    goes down to -1 at 180 degrees,
    comes back up to zero at 270 degrees,
    reaches 1 again at 360 degrees.

    Therefore there is an interval from zero to where cos(x) is \frac{\sqrt{3}}{2}

    where cos(x)>\frac{\sqrt{3}}{2}

    and similar at the other end of the period.
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  3. #3
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    yes, that is the graph i plotted. I think I dont understand how to figure out the interval where  cos(x)>\frac{\sqrt{3}}{2} . I can see the area where it is greater than  \frac{\sqrt{3}}{2} , but i dont know how to define it.

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  4. #4
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    Oh! I think I got it.

    Is the answer  [0,\frac{\pi}{6}) U (\frac{11\pi}{6},2\pi]
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  5. #5
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    Note that you need to solve \cos x-\cos \frac{\pi }{6}=2\sin \left( \frac{6x+\pi }{12} \right)\sin \left( \frac{\pi -6x}{12} \right)>0.
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  6. #6
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    you've got it, discobob
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